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In problem 3 for the following given equation resulted solution which satisfies CNF definition.x <-> (y1 & y2)[x -> (y1 & y2)] & [(y1 & y2) -> x][ !x | (y1 & y2) ] & [ !(y1 & y2) | x ](~x | y1) & (~x | y2) & (!y1 | !y2 | x)But the system shows it as incorrect. What should be modified in this case? Isn't it satisfies CNF definition?
Update: actually submitted (!x | y1) & (!x | y2) & (x | !y1 | !y2)
Let's go through it step by step:
x ↔ (y1 ∧ y2) // the formula[x →(y1 ∧ y2) ] ∧ [(y1 ∧ y2) → x] // display ↔ as a conjunction of two implications[¬x ∨ (y1 ∧ y2) ] ∧ [¬(y1 ∧ y2) ∨ x] // remove the implication syntactic sugar(¬x ∨ y1) ∧ (¬x ∨ y2) ∧ (¬y1 ∨ ¬y2 ∨ x) // multiply out (left), apply de-morgan (right)
So your formula does make sense to me.
Make sure that you don't confuse ^/⊕/xor for &/∧/and as shown here: https://q2a.cs.uni-kl.de/469/exercise-sheet-1-task-3-e-broken